What is the derivative of (1+8/x ^x) ?

The derivative of (1+8/x ^x) is (ln(1+8/x)-8/(x+8))(1+8/x)^x

What is the first derivative of (1+8/x ^x) ?

The first derivative of (1+8/x ^x) is (ln(1+8/x)-8/(x+8))(1+8/x)^x

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\frac{d}{d x} ((1 + \frac{8}{x})^{x})

(ln (1 + \frac{8}{x}) - \frac{8}{x + 8}) (1 + \frac{8}{x})^{x}

Solution steps

\frac{d}{d x} ((1 + \frac{8}{x})^{x})

Apply exponent rule:

a^{b} = e^{b l n (a)}

(1 + \frac{8}{x})^{x} = e^{x l n (1 + \frac{8}{x})}

= \frac{d}{d x} (e^{x l n (1 + \frac{8}{x})})

Apply the chain rule:

e^{x l n (1 + \frac{8}{x})} \frac{d}{d x} (x ln (1 + \frac{8}{x}))

= e^{x l n (1 + \frac{8}{x})} \frac{d}{d x} (x ln (1 + \frac{8}{x}))

\frac{d}{d x} (x ln (1 + \frac{8}{x})) = ln (1 + \frac{8}{x}) - \frac{8}{x + 8}

= e^{x l n (1 + \frac{8}{x})} (ln (1 + \frac{8}{x}) - \frac{8}{x + 8})

e^{x l n (1 + \frac{8}{x})} = (1 + \frac{8}{x})^{x}

= (ln (1 + \frac{8}{x}) - \frac{8}{x + 8}) (1 + \frac{8}{x})^{x}

\frac{d}{d x} (x + 1 - x)

\frac{d y}{d t} + e^{t} y = - 4 e^{t}

x \to 0 + lim (\frac{- 4 x ^{2} - 6 x}{x})

f (x) = \frac{3}{4} x^{8}

\frac{d}{d y} (3 x (5 + y)^{- 1})

What is the derivative of (1+8/x ^x) ?
The derivative of (1+8/x ^x) is (ln(1+8/x)-8/(x+8))(1+8/x)^x
What is the first derivative of (1+8/x ^x) ?
The first derivative of (1+8/x ^x) is (ln(1+8/x)-8/(x+8))(1+8/x)^x